Maths&physics
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any ideas how I'd tackle this?
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do I use this formula and make a = 0?
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mqb2766
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(Original post by Maths&physics)
do I use this formula and make a = 0?
Presume so. The initial conditions give you the first two terms. The ode (rearranged) would give you the third term (t^2) and you could differentiate the ode again, to get the last (t^3) term.
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RDKGames
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(Original post by Maths&physics)
any ideas how I'd tackle this?
You seek a truncated expansion of x(t), where it can be expanded in powers of t as:

x(t) = x(0) + x'(0)t + \dfrac{x''(0)}{2}t^2 + \dfrac{x'''(0)}{6}t^3 + O(t^4)

t=0 implies x(0) = 0 and x'(0) = \dfrac{1}{2}, this means that

x(t) = 0 + \dfrac{1}{2}t + \dfrac{x''(0)}{2}t^2 + \dfrac{x'''(0)}{6}t^3 + O(t^4)

To get the fourth term, you can differentiate the ODE and see what x'''(0) is. The second term should be fairly obvious.
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(Original post by RDKGames)
You seek a truncated expansion of x(t), where it can be expanded in powers of t as:

x(t) = x(0) + x'(0)t + \dfrac{x''(0)}{2}t^2 + \dfrac{x'''(0)}{6}t^3 + O(t^4)

t=0 implies x(0) = 0 and x'(0) = \dfrac{1}{2}, this means that

x(t) = 0 + \dfrac{1}{2}t + \dfrac{x''(0)}{2}t^2 + \dfrac{x'''(0)}{6}t^3 + O(t^4)

To get the fourth term, you can differentiate the ODE and see what x'''(0) is. The second term should be fairly obvious.
(Original post by mqb2766)
Presume so. The initial conditions give you the first two terms. The ode (rearranged) would give you the third term (t^2) and you could differentiate the ode again, to get the last (t^3) term.
what is ODE?
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mqb2766
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Sorry, the differential equation. Just differentiate it with respect to x again.
ODE is Ordinary Differential Equation - you don't need to know the term, but can look it up if you want.

(Original post by Maths&physics)
what is ODE?
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RDKGames
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(Original post by Maths&physics)
what is ODE?
Oh dear...

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation


I.e. the equation with the derivative that they give you!
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(Original post by RDKGames)
Oh dear...

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation


I.e. the equation with the derivative that they give you!
is that a term in the specification because I've never heard it before. :confused:
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mqb2766
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(Original post by Maths&physics)
is that a term in the specification because I've never heard it before. :confused:
No.
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RDKGames
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(Original post by Maths&physics)
is that a term in the specification because I've never heard it before. :confused:
I don't know... check your spec online and CTRL+F for "ODE" to see if it appears or not...


But TBH, I would be very surprised (and disappointed) if the resources you use for this theory do not abbreviate the topic to ODE because it's an extremely standard acronym.
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(Original post by mqb2766)
No.
(Original post by RDKGames)
Oh dear...

That is really not a question you want to be asking at this stage...

ODE = Ordinary Differential Equation


I.e. the equation with the derivative that they give you!
I got for the third term... (d^3)y/d(t^3) = + (dx/dt)[1-sin(x)]dx/dt = 0

I've got values for x but what would dx/dt =?
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mqb2766
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(Original post by Maths&physics)
I got for the third term... (d^3)y/d(t^2) = + (dx/dt)[1-sin(x)]dx/dt = 0

I've got values for x but what would dx/dt =?
Its evaulated at t=0. You're given the value in the initial condition.
Edit that expression has gone a bit wonky?
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(Original post by Maths&physics)
I got for the third term... (d^3)y/d(t^2) = + (dx/dt)[1-sin(x)]dx/dt = 0

I've got values for x but what would dx/dt =?
That equation looks dodgy... how did you arrive at that??
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(Original post by mqb2766)
Its evaulated at t=0. You're given the value in the initial condition.
Edit that expression has gone a bit wonky?
(Original post by RDKGames)
That equation looks dodgy... how did you arrive at that??
I meant: (d^3)y/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0

????

ok, what do I do next? I don't understand how to get dx/dt?

I sub x = 0, and I get: (d^3)y/d(t^3) = -(dx/dt)
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RDKGames
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(Original post by Maths&physics)
I meant: (d^3)y/d(t^3) = + (dx/dt)[1-sin(x)]dx/dt = 0

????

ok, what do I do next?
That's not right.

(a) what are you actually doing to this equation...?

(b) why do you have =0 there...?
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#16
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(Original post by RDKGames)
That's not right.

(a) what are you actually doing to this equation...?

(b) why do you have =0 there...?
(d^3)y/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0

(d^3)y/d(t^3) = -(dx/dt)
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RDKGames
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(Original post by Maths&physics)
(d^3)y/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0

(d^3)y/d(t^3) = -(dx/dt)
Nope... [second line is fine tho...]
You should really lay it out properly.

\dfrac{d^2 x}{dt^2} + x + \cos (x) = 0

\implies \dfrac{d^3 x}{dt^3} + \dfrac{dx}{dt} - \dfrac{dx}{dt} \sin (x) = 0

When t=0, we get:

x'''(0) + x'(0) - x'(0)\sin[x(0)] = 0

You know every single value here apart from x'''(0) ... so rearrange for it.
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#18
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(Original post by RDKGames)
Nope...


\dfrac{d^2 x}{dt^2} + x + \cos (x) = 0

\implies \dfrac{d^3 x}{dt^3} + \dfrac{dx}{dt} - \dfrac{dx}{dt} \sin (x) = 0

When t=0, we get:

x'''(0) + x'(0) - x'(0)\sin[x(0)] = 0

You know every single value here apart from x'''(0) ... so rearrange for it.
\implies \dfrac{d^3 x}{dt^3} + \dfrac{dx}{dt} - \dfrac{dx}{dt} \sin (x) = 0

that's what I meant!

(d^3)x/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0
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(Original post by Maths&physics)
\implies \dfrac{d^3 x}{dt^3} + \dfrac{dx}{dt} - \dfrac{dx}{dt} \sin (x) = 0

that's what I meant!

(d^3)x/d(t^3) + (dx/dt)[1-sin(x)]dx/dt = 0
(Original post by mqb2766)
Its evaulated at t=0. You're given the value in the initial condition.
Edit that expression has gone a bit wonky?
I'm so sorry, I confused my notations with the tutorial I watching.
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